Answer:tbh, that's impossible
Step-by-step explanation:
Answer:
<h2>x = 3 and x = 4 → (3, 0) and (4, 0)</h2>
Step-by-step explanation:

The answer would be A = 54raiz (3) + 18raiz (91)
Formula:
A = Ab + Al Where, Ab=base area A= lateral area
The area of the base is: Ab = (3/2) * (L ^ 2) * (root (3)) Where, L= side of the hexagon. Substitute: Ab = (3/2) * (6 ^ 2) * (root (3)) Ab = (3/2) * (36) * (root (3)) Ab = 54raiz (3)
The lateral area is: Al = (6) * (1/2) * (b) * (h) Where, b= base of the triangle h= height of the triangle Substitute: Al = (6) * (1/2) * (6) * (root ((8) ^ 2 + ((root (3) / 2) * (6)) ^ 2)) Al = 18 * (root (64 + 27)) Al = 18raiz (91)
The total area is: A = 54raiz (3) + 18raiz (91)
Answer:
C
Step-by-step explanation:
It usually works best to use the polynomial with fewer terms as the multiplier. A row of partial products is written for each term of the multiplier, so the fewer terms will result in fewer rows of partial products.
In order to keep like terms together, it is preferable to allocate a separate column of the multiplication tableau to each power of the operands or product. This means we want to make note of the fact that the cubic multiplicand has a coefficient of 0 for its x^2 term.
The best setup is the one shown in the attachment.

Let the capacity of bus be x students
And van be y students, now ;
From the given statements we get two equations ~


multiply the equation (2) with 2 [ it won't change the values ]


Now, deduct equation (1) from equation (3)




Therefore each bus can carry (x) = 45 students
Now, plug the value of x in equation (1) to find y ~







Hence, each van can carry (y) = 17 students in total.